Finding the Concentrations of Ions Left in Solution after Selective Precipitation
You add potassium hydroxide to the solution in Example 17.13. When the $[OH^-]$ reaches $1.9 × 10^{-6} \ M$ (as you just calculated), magnesium hydroxide begins to precipitate out of solution. As you continue to add KOH, the magnesium hydroxide continues to precipitate. At some point, the $[OH^-]$ becomes high enough to begin to precipitate the calcium ions as well. What is the concentration of $Mg^{2+}$ when $Ca^{2+}$ begins to precipitate?

Question

Finding the Concentrations of Ions Left in Solution after Selective Precipitation
You add potassium hydroxide to the solution in Example 17.13. When the [latex][OH^-][/latex] reaches [latex]1.9 × 10^{-6} \ M[/latex] (as you just calculated), magnesium hydroxide begins to precipitate out of solution. As you continue to add KOH, the magnesium hydroxide continues to precipitate. At some point, the [latex][OH^-][/latex] becomes high enough to begin to precipitate the calcium ions as well. What is the concentration of [latex]Mg^{2+}[/latex] when [latex]Ca^{2+}[/latex] begins to precipitate?

Accepted Answer

First, calculate the [latex]OH^-[/latex] concentration at which [latex]Ca^{2+}[/latex] begins to precipitate by writing the expression for Q for calcium hydroxide and substituting the concentration of [latex]Ca^{2+}[/latex] from Example 17.13.

[latex] \mathcal{Q}= [Ca^{2+}][OH^-]^2[/latex]
[latex]= (0.011)[OH^-]^2[/latex]

Set the expression for Q equal to the value of [latex]K_{sp}[/latex] for calcium hydroxide and solve for [latex[OH^-][/latex]. [latex]Ca(OH)_2[/latex] precipitates above this concentration.

When [latex] \mathcal{Q}=K_{sp}[/latex]

[latex](0.011)[OH^-]^2 = K_sp = 4.68 × 10^{-6}[/latex]

[latex][OH^-]^2=\frac{4.68 imes 10^{-6}}{0.011} [/latex]

[latex][OH^-] = 2.\underline{0} 6 imes 10^{-2}[/latex] M

Find the concentration of [latex]Mg^{2+}[/latex] when [latex]OH^-[/latex] reaches the concentration you just calculated by writing the expression for Q for magnesium hydroxide and substituting the concentration of [latex]OH^-[/latex] that you just calculated. Then set the expression for Q equal to the value of [latex]K_{sp}[/latex] for magnesium hydroxide and solve for [latex][Mg^{2+}][/latex]. This is the concentration of [latex]Mg^{2+}[/latex] that remains when [latex]Ca(OH)_2[/latex] begins to precipitate.

[latex] \mathcal{Q}= [Mg^{2+}][OH^-]^2[/latex]

[latex]= [Mg^{2+}](2.\underline{0} 6 imes 10^{-2})^2[/latex]

When [latex] \mathcal{Q}=K_{sp}[/latex]

[latex][Mg^{2+}](2.\underline{0} 6 imes 10^{-2})^2=K_{sp}=2.06 imes 10^{-13}[/latex]

[latex][Mg^{2+}]=\frac{2.06 imes 10^{-13}}{(2.\underline{0} 6 imes 10^{-2})^2} [/latex]

[latex][Mg^{2+}]= 4.9 imes 10^{-10}[/latex] M

As you can see from the results, the selective precipitation worked very well. The concentration of [latex]Mg^{2+}[/latex] dropped from 0.059 M to [latex]4.9 imes 10^{-10}[/latex] M before any calcium began to precipitate, which means that 99.99% of the magnesium separated out of the solution

Question 17.14: Finding the Concentrations of Ions Left in Solution after Se...

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